Assignment 27

course Mth 151

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027. `query 27

College Algebra

04-20-2007

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21:23:22

5.5.6 Give the approximate value of Golden Ratio to thousandth and show how you obtained your result.

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RESPONSE -->

[1+ sqrt(5)] / 2 = [1+2.2361] / 2 = 3.2361 / 2 = 1.618

confidence assessment: 3

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21:23:42

** The Golden Ratio is [ 1+`sqrt(5) ] /2

[ 1+`sqrt(5) ] /2=[ 1+2.2361 ] /2=3.2361/2=1.618 **

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RESPONSE -->

Ok.

self critique assessment: 3

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21:27:37

5.5.12 2^3 + 1^3 - 1^3 = 8; 3^3 + 2^3 - 1^3 = 34; 5^3 + 3^3 - 2^3 = 144; 8^3 + 5^3 - 3^3 = 610.

What are the next two equations in this sequence?

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RESPONSE -->

I'm not sure.

confidence assessment: 0

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21:28:42

** The numbers 1, 1, 2, 3, 5, 8, ... are the Fibonacci numbers f(1), f(2), ... The left-hand sides are

f(3)^3 + f(2)^3 - f(1)^3,

f(4)^3 + f(3)^3 - f(2)^3,

f(5)^3 + f(4)^3 - f(3)^3,

f(6)^3 + f(5)^3 - f(4)^3 etc..

The right-hand sides are f(5) = 8, f(8) = 34, f(11) = 144, f(14) = 610. So the equations are

f(3)^3 + f(2)^3 - f(1)^3 = f(5)

f(4)^3 + f(3)^3 - f(2)^3 = f(8)

f(5)^3 + f(4)^3 - f(3)^3 = f(11)

f(6)^3 + f(5)^3 - f(4)^3 = f(14)

etc..

The next equation would be

f(7)^3 + f(6)^3 - f(5)^3 = f(17).

Substituting f(7) = 13, f(6) = 8 and f(5) = 5 we get

13^3 + 8^3 - 5^3 = f(17). The left-hand side gives us result 2584, which is indeed f(17), so the pattern is verified in this instance. **

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RESPONSE -->

Oh, ok I didn't recognize the Fibonacci numbers.

self critique assessment: 2

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21:33:39

5.5.18 show whether F(p+1) or F(p-1) is divisible by p.

Give your solution to this problem.

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RESPONSE -->

f(p-1) = f(2) = 1

f(p+1) = f(4) = 3 ... Can be divided by P ... 3

f(p-1) = f(6) = 8

f(p+1) = f(8) = 21 ... Can be divided by P ... 7

f(p-1) = f(10) = 55

f(p+1) = f(12) = 144 ... Can be divded by P ... 11

confidence assessment: 3

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21:34:05

** For p=3 we get f(p-1) = f(2) = 1 and f(p+1) = f(4)= 3; f(p+1) = f(4) = 3 is divisible by p, which is 3 So the statement is true for p = 3.

For p=7 we get f(p-1) = f(6) = 8 and f(p+1) = f(8) = 21; f(p+1) = 21 is divisible by p = 7. So the statement is true for p = 7.

For p = 11 we get f(p-1) = f(10)= 55 and f(p+1) = f(12) = 144. f(p-1) = 55 is divisible by p = 11. So the statement is true for p = 11.

So the conjecture is true for p=3, p=7 and p=11.**

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RESPONSE -->

Ok.

self critique assessment: 3

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21:38:06

5.5.24 Lucas sequence: L2 + L4; L2 + L4 + L6; etc.. Give your solution to this problem as stated in your text.

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RESPONSE -->

L2 + L4 = 3 + 7 = 10

L2 + L4 + L6 = 3 + 7 + 18 = 28

L2 + L4 + L6 + L8 = 3 + 7 + 18 + 47 = 75

L2 + L4 + L6 + L8 + L10 = 3 + 7 + 18 + 47 + 123 = 198

confidence assessment: 3

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21:38:43

** The Lucas sequence is 1 3 4 7 11 18 29 47 76 123 199 etc.

L2 + L4 = 3 + 7 = 10;

L2 + L4 + L6 = 3 + 7 + 18 = 28;

L2 + L4 + L6 + L8 = 3 + 7 + 18 + 47 = 75, and

L2 + L4 + L6 + L8 + L10 = 3 + 7 + 18 + 47 + 123 = 198.

Note that 10 is 1 less than 11, which is L5; 27 is 1 less than 28, which is L7; and 198 is 1 less than 199, which is L9.

So L2 + L4 = L5 - 1, L2 + L4 + L6 = L7 - 1, etc..

So we can conjecture that the sum of a series of all evenly indexed members of the Lucas sequence, starting with index 2 and ending with index 2n, is 1 less than member 2n+1 of the sequence. **

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RESPONSE -->

Ok.

self critique assessment: 3

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Assignment 27

Very good. Let me know if you have any questions.